Let $A = \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix}$, where $0 \leq \theta \leq 2\pi$. Then which of the following are true?

(A) $|A| = 2 + 2 \sin^{2} \theta$

(B) $|A| = 2 + \sin^{2} \theta$

(C) minimum value of $|A|$ is $1$

(D) maximum value of $|A|$ is $4$

Choose the correct answer from the options given below:

We need to find the determinant $|A|$ of the given $3 \times 3$ matrix and then determine its minimum and maximum values.

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For the matrix $A = \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix}$, we'll expand along Row 1:

$|A| = 1 \cdot \begin{vmatrix} 1 & \sin \theta \\ -\sin \theta & 1 \end{vmatrix} - \sin \theta \cdot \begin{vmatrix} -\sin \theta & \sin \theta \\ -1 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} -\sin \theta & 1 \\ -1 & -\sin \theta \end{vmatrix}$

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Calculating each $2 \times 2$ determinant:

$\begin{vmatrix} 1 & \sin \theta \\ -\sin \theta & 1 \end{vmatrix} = 1 + \sin^2 \theta$

$\begin{vmatrix} -\sin \theta & \sin \theta \\ -1 & 1 \end{vmatrix} = -\sin \theta + \sin \theta = 0$

$\begin{vmatrix} -\sin \theta & 1 \\ -1 & -\sin \theta \end{vmatrix} = \sin^2 \theta + 1$

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Substituting back:

$|A| = 1(1 + \sin^2 \theta) - \sin \theta(0) + 1(\sin^2 \theta + 1)$

$|A| = 1 + \sin^2 \theta + \sin^2 \theta + 1$

$|A| = 2 + 2\sin^2 \theta$

This confirms option (A) is correct.

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Since $0 \leq \sin^2 \theta \leq 1$:

Minimum value occurs when $\sin^2 \theta = 0$:

$|A|_{\text{min}} = 2 + 2(0) = 2$

Maximum value occurs when $\sin^2 \theta = 1$:

$|A|_{\text{max}} = 2 + 2(1) = 4$

This confirms option (D) is correct.

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Options (A) and (D) are correct.